A) \[\left( 0,\frac{\pi }{2} \right)\]
B) \[\left[ \underset{{}}{\mathop{0,}}\, \right.\left. \frac{\pi }{2} \right)\]
C) \[\left( \left. 0,\frac{\pi }{2} \right] \right.\]
D) \[\left[ 0,\,\frac{\pi }{2} \right]\]
Correct Answer: B
Solution :
We have the function \[y={{\sin }^{-1}}\left( \frac{{{x}^{2}}}{1+{{x}^{2}}} \right)\] For y to be defined \[\left| \frac{{{x}^{2}}}{1+{{x}^{2}}} \right|<1\] which is true for all \[x\in R\]. Now, \[y={{\sin }^{-1}}\left( \frac{{{x}^{2}}}{1+{{x}^{2}}} \right)\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{1+{{x}^{2}}}=\sin y\] \[\Rightarrow \] \[x=\sqrt{\frac{\sin y}{1-\sin y}}\] For the existance of x \[\sin y\ge 0\]and \[1-\sin y>0\] \[\Rightarrow \] \[0\le \sin y<1\] \[\Rightarrow \] \[0\le y<\frac{\pi }{2}\] Thus, range of the given function is\[\left[ 0,\,\frac{\pi }{2} \right)\].
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